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Finding the focus of a parabola given an equation

  1. Mar 26, 2016 #1
    1. The problem statement, all variables and given/known data
    Alright, so the equation of a parabola is [itex]y = 1/4p*x^2[/itex], P being either an x or y value, and the other x or y being zero. Let's say that [itex]x^2 = 16y[/itex]. If you divide both sides by 16, you get [itex]y = x^2/16[/itex], which can be simplified to [itex]y = 1/16*x^2[/itex]. This is in the format of a parabola, so finding p is simple. 16 is the product of 4p, so [itex]4p=16[/itex]. Divide both sides by 4 and p=4. So the focus is at (0,4).

    This seems simple to me, until you get equations like this: [itex]3x^2 + 4y = 0[/itex]. I can't seem to get this in the form of a parabola. I got [itex]-3/4 *x^2=y[/itex]. This wouldn't help me find the focus, due to the fact that it's not in that form. How exactly would you go about finding the focus in cases like this? Thanks for any help!
     
    Last edited by a moderator: Mar 26, 2016
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  3. Mar 26, 2016 #2

    SammyS

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    Hello BlueQuark. Welcome to PF !

    You need parentheses around the 4p since both are in the denominator. You are using LaTeX, so it is easy to use "\frac" to write : ##\ y = \frac{1}{4p}x^2\ ##.

    Writing the equation of the parabola as ##\ 4py = x^2\ ## may prove to be even handier for solving your problem.
     
  4. Mar 26, 2016 #3

    Math_QED

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    4y + 3x^2 = 0
    <=> y = -3/4 x^2
    <=> y = -1/(4/3)x^2

    I don't know whether this has answered your question.
     
  5. Mar 26, 2016 #4

    HallsofIvy

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    As SammyS said, that should be [itex]y= 1/(4p)x^2[/itex] or [itex]y= x^2/(4p)[/itex]

    I really don't know what that means. What does it mean to say that p is "either an x or y value". What do you mean by an "x value" or a "y value"? And what "other" x or y do you mean? p is, of course, the x coordinate of the focus- in this situation, the focus is at (0, p). Is that what you mean?

    Surely you know that [itex]-3/4= \frac{1}{-\frac{4}{3}}[/itex]? [itex]p= -\frac{4}{3}[/itex] and the focus is at [itex]\left(0, -\frac{4}{3}\right)[/itex].
     
  6. Mar 27, 2016 #5

    Math_QED

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    Yes, [itex]-3/4= \frac{1}{-\frac{4}{3}}[/itex]

    And, y = -x^2/(4p) is a valid equation for a parabola.
    P = -1/3, not -4/3 so the focus is at (0, -1/3)
     
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